Question

A company services home air conditioners. It is known that times for service calls follow a normal distribution with a mean of 75 minutes and a standard deviation of 15 minutes. A random sample of twelve service calls is taken. What is the probability that exactly eight of them take more than 93.6 minutes

Answer 1

Answer:

The probability that exactly eight of them take more than 93.6 minutes is 5.6015 $\times 10^{-6}$ .

Step-by-step explanation:

We are given that it is known that times for service calls follow a normal distribution with a mean of 75 minutes and a standard deviation of 15 minutes.

A random sample of twelve service calls is taken.

So, firstly we will find the probability that service calls take more than 93.6 minutes.

Let X = times for service calls.

So, X ~ Normal($\mu=75,\sigma^{2} =15^{2}$)

The z-score probability distribution for the normal distribution is given by;

                              Z  =  $\frac{X-\mu}{\sigma}$  ~ N(0,1)

where, $\mu$ = mean time = 75 minutes

           $\sigma$ = standard deviation = 15 minutes

Now, the probability that service calls take more than 93.6 minutes is given by = P(X > 93.6 minutes)

       P(X > 93.6 min) = P( $\frac{X-\mu}{\sigma}$ > $\frac{93.6-75}{15}$ ) = P(Z > 1.24) = 1 - P(Z $\leq$ 1.24)

                                                                = 1 - 0.8925 = 0.1075

The above probability is calculated by looking at the value of x = 1.24 in the z table which has an area of 0.8925.

Now, we will use the binomial distribution to find the probability that exactly eight of them take more than 93.6 minutes, that is;

$P(Y = y) = \binom{n}{r}\times p^{r} \times (1-p)^{n-r} ; y = 0,1,2,3,.........$

where, n = number of trials (samples) taken = 12 service calls

            r = number of success = exactly 8

            p = probability of success which in our question is probability that

                   it takes more than 93.6 minutes, i.e. p = 0.1075.

Let Y = Number of service calls which takes more than 93.6 minutes

So, Y ~ Binom(n = 12, p = 0.1075)

Now, the probability that exactly eight of them take more than 93.6 minutes is given by = P(Y = 8)

               P(Y = 8)  =  $\binom{12}{8}\times 0.1075^{8} \times (1-0.1075)^{12-8}$

                             =  $495 \times 0.1075^{8} \times 0.8925^{4}$

                             =  5.6015 $\times 10^{-6}$ .